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EXPLANATION OF SILVER IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 5, 2015 Silver is a chemical element with symbol Ag and atomic number 47. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Silver including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s1 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of silver (from (E1 to E3 ) are the following: E1 = 7.58 , E2 = 21.49, and E3 = 34.83 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper of 2008. EXPLANATION OF E1 = 7.58 eV = -E(5s1) ''' Here the E(5s1) represents the binding energy of 5s1. The charges (-46e) of (1s22s22p63s23p63d10 4s2 4p64d10 ) screen the nuclear charge (+47e) and for a perfect screening we would have ζ = 1. However the electron of 5s1 penetrates the 4d10 and leads to the deformations of electron clouds. Thus ζ > 1. Since 5s1 consists of one electron, we apply the Bohr formula to write E(5s1) = 7.58 eV = - (-13.6057)ζ2/n2 Therefore E1 = 7.58 eV = (13.6057)ζ2 / 52 Then we get ζ = 3.73 > 1 . '''EXPLANATION OF Ε2 = 21.49 eV = -E(4d10) + E(4d9 ) Here the E(4d10) represents the binding energy of the 10 electrons (4d10), while the E(4d9) represents the binding energy of 9 electrons (4d9) .The charges (-36e) of (1s22s22p63s23p63d104s24p6 ) screen the nuclear charge (+47e) and for a perfect screening we would have an effective ζ = 11. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p6 but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 11. Note that 4d10 consists of five pairs (10 electrons with opposite spin) Thus for the five pairs of opposite spin we apply my formula of 2008. Under this condition one may write -E(4d10) = -5ζ2 + (16.95)ζ - 4.1 / n2 On the other hand, the 4d9 consists of four pairs (8 electrons with opposite spin ) whose the binding energy is given by applying my formula of 2008, and of one electron whose the binding energy is given by applying the Bohr formula. So we may write E(4d9 ) = 4+ (16.95)ζ - 4.1 / n2 + (13.6057)ζ2 / n2 Therefore Ε2 = 21.49 eV = -E(4d10) + E(4d9 ) = - (16.95)ζ + 4.1 / n2 It is of interest to note that in this sub- shell of 4d the experiments of ionizations showed that 4 < n < 5. (See my paper EXPLANATION OF TECHNETIUM IONIZATIONS). Thus using n = 4 one writes (13.6057)ζ2 - (16.95)ζ - 339.74 = 0 and solving for ζ we get ζ = 5.66 < 11 Whereas using n = 5 one writes (13.6057)ζ2 - (16.95)ζ - 533.15 = 0 and solving for ζ we get ζ = 6.91 < 11 . Of course the two electrons of opposite spin (say the 4dx2 ) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy. This situation indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of the invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' EXPLANATION OF E3 = 34.83 eV = -E(4d9) + E(4d8) Here the E(4d9) represents the binding energy of the 9 electrons (4d9), while the E(4d8) represents the binding energy of 8 electrons (4d8) . As in the case of E2 the charges (-36e) of (1s22s22p63s23p63d104s24p6 ) screen the nuclear charge (+47e) and for a perfect screening we would have the same effective ζ = 11. However, in this case, the experiments of ionizations show that each pair of 4d repels not only the electrons of 4p6 but also the electrons of 4d. Surprisingly such repulsions occur in such a way that ζ < 11 . Note that the 4d9 consists of four pairs (8 electrons with opposite spin ) whose the binding energy is given by applying my formula of 2008, and of one electron whose the binding energy is given by applying the Bohr formula. So we may write -E(4d9 ) = - 4+ (16.95)ζ - 4.1 / n2 - (13.6057)ζ2 / n2 On the other hand the 4d8 consists of 3 pairs (6 electrons with opposite spin), whose the binding energy is given by applying my formula of 2008, and of 2 electrons with parallel spin, whose the binding energy is given by applying the Bohr formula. Under this condition one may write E(4d8) = 3+ (16.95)ζ - 4.1 / n2 + 2(-13.6057)ζ2 /n2 Therefore Ε3 = 34.83 eV = - E(4d9) + E(4d8 ) = - (16.95)ζ + 4.1 / n2 As in the case of E2 in this sub- shell of 4d the experiments of ionizations showed that 4 < n < 5. Thus using n = 4 one writes (13.6057)ζ2 - (16.95)ζ - 553.18 = 0 and solving for ζ we get ζ = 7.03 < 11 Whereas using n = 5 one writes (13.6057)ζ2 - (16.95)ζ - 866.65 = 0 and solving for ζ we get ζ = 8.63 < 11. Category:Fundamental physics concepts